Intrinsic proof that sprays induce involutions

Question

Let $M$ be a smooth manifold. Let $V$ be the canonical vector field on $T M$ (also called the Liouville vector field), which if $(x, y)$ are local coordinates on $T M$ is defined by $V = y^i \frac{\partial }{\partial y^i}$. Also denote by $J : TTM \to TTM$ the canonical endomorphism of $TTM$ (also called an almost tangent structure), which in local coordinates is given by $J(a^i \frac{\partial }{\partial x^i} + b^i \frac{\partial }{\partial y^i}) = a^i \frac{\partial }{\partial y^i}$.

These definitions immediately imply some basic facts:

• $\mathcal{L}_V J = -J$,
• $J^2 = 0$ (in fact $\operatorname{im} J = \operatorname{ker} J$), and
• $[JX, JY] = J[JX, Y] + J[X, JY]$.

Now let $S$ be a spray on $M$, which means that $S$ is a vector field on $TM$ such that (1) $J S = V$ and (2) $[V, S] = S$. In this situation, the Lie derivative $\mathcal{L}_S J$ is an involution on $TTX$, which is intimately connected to the affine connection $S$ induces on $M$. (Namely $S$ defines an Ehresmann connection from the horizontal projection $h : TTM \to TTM$ given by $\frac{1}{2}(1 - \mathcal{L}_S J)$.)

Is it possible to prove that $(\mathcal{L}_S J)^2 = 1$ intrinsically (without calculating in coordinates) using these facts (or additional ones)? I can quickly show $(\mathcal{L}_S J)^2 = 1$ in coordinates, and coordinate-freely after some manipulations I can obtain $$ (\mathcal{L}_S J)^2(X) = -[S, JX] - J[S, X] - J[S, [S, JX]]. $$ Is it possible to make further progress algebraically with these given facts? If not, is there an additional algebraic identity which will? Is there a moral reason why a coordinate-free proof along these lines is hard to come by?

Answer 1

It turns out that this is just a linear algebra problem. We don't even need the condition $[V, S] = S$ (such $S$ is then just called a semispray).

First observe that since $(\mathcal{L}_S J)X = [S, JX] - J[S, X]$ we easily have $$ J(\mathcal{L}_S J)X = J[S, JX] = [V, JX] - J[V, X] = (\mathcal{L}_V J)X = -JX, $$ and $(\mathcal{L}_S J) J X = -J[S, JX] = JX$ similarly. Now we just appeal to the following lemma.

Lemma. Let $J$ and $T$ be endomorphisms of a finite dimensional vector space $V$. If $\operatorname{im} J = \operatorname{ker} J$, $JT = -J$, and $TJ = J$, then $T^2 = 1$.

Proof. The condition $TJ = J$ implies that the $1$-eigenspace of $T$ is at least $\dim \ker J$ i.e $\frac{1}{2} \dim V$-dimensional. The condition $JT = -J$ i.e. $J (T + 1) = 0$ implies that $\dim \operatorname{im} (T + 1) \leq \frac{1}{2} \dim V$. Hence the $-1$-eigenspace of $T$ is at least $\frac{1}{2} \dim V$-dimensional as well. It follows that $T^2$ is diagonalizable with sole eigenvalue $1$, i.e. $T^2 = 1$.

Answer 2

Not sure this is an answer, but it is too long for a comment. One of the reasons why the proof might not be shorter is that the intrinsic definition of the canonical endomorphism first projects and then lifts: $J(v)=(\pi_*(v))^V$, with $(\cdot)^V$ the vertical lift and $\pi$ the tangent bundle projection. While easy in coordinates, working with the above may not be straightforward.

If you are familiar with vertical and complete lifts, there is an alternative approach. You can then show that $(\mathcal{L}_SJ)(X^V)=X^V$ and $(\mathcal{L}_SJ)^2(X^C)=X^C$. The first one is easy, the second one involves more work. From here, since $\{X^V,X^C\}$ span $TTM$ you are done. You can find this computation, or similar ones, in a few papers. One of the first ones I am aware of is the following:

Crampin, M. - Tangent bundle geometry for Lagrangian dynamics. J. Phys. A 16, 3755-3772 (1983).